Question

a taxi changes Rs 8 for first one km. and Rs 5/km for rest distance. then find the eqation​

Answer 1

Answer:

Given, Total distance covered=x km=1+(x-1)km

Fare For first kilometre= R8

2. 4.00

Fare for subsequent distance= 5 per km

Fare for next(x-1)km= 5(x-1)

A.T.Q

Total fare=y

8+5(x-1)=y

8+5x-5=y

5x-y+3=0

Which is the required linear equation.

It can also be written as y= 5x+3

When X = 0,then Y = 3,When x=1, then y= 5+3=8

When x=2, then y= 10+3=13

[Table & graph are on the attachment]

Now plot the points A(0,3), B(1,8), C

2,13) on the graph paper and join them to form a line BC, which represents the required graph of linear equation.

Answer 2

Step-by-step explanation:

let variable be y

8 +5y = whatever given

whatever value of y will come add 1 to that to get answer